Question

A 5 -m-diameter spherical tank is filled with liquid oxygen $\left(\rho=1141 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1.71 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\( at \)-184^{\circ} \mathrm{C}$. It is observed that the temperature of oxygen increases to \(-183^{\circ} \mathrm{C}\) in a 144-hour period. The average rate of heat transfer to the tank is (a) \(124 \mathrm{~W}\) (b) \(185 \mathrm{~W}\) (c) \(246 \mathrm{~W}\) (d) \(348 \mathrm{~W}\) (e) \(421 \mathrm{~W}\)

Short answer

Based on the given scenario, calculate the average rate of heat transfer to the spherical tank filled with liquid oxygen. Answer: (c) 246 W

Step-by-step solution

Determine the volume of the spherical tank.

To calculate the volume of the spherical tank, we can use the formula for the volume of a sphere, given by: \(V = \frac{4}{3}\pi r^3\) Given the diameter of the tank, 5 meters, we can find the radius as: \(r = \frac{5}{2} = 2.5 \mathrm{m}\) Now, calculate the volume \(V\): \(V = \frac{4}{3}\pi (2.5)^3 = 65.45 \mathrm{m^3}\)

Calculate the mass of liquid oxygen in the tank.

Using the density of liquid oxygen, we can find the mass of the liquid oxygen in the tank. The formula to find the mass is: \(m = \rho V\) Let's plug in the values we have: \(m = 1141 \mathrm{kg/m^3} \times 65.45 \mathrm{m^3} = 74619.45 \mathrm{kg}\)

Calculate the total heat transferred.

The total heat transferred can be found using the formula: \(Q = m c_p \Delta T\) We know the values of mass (\(m\)), specific heat capacity (\(c_p\)), and temperature difference (\(\Delta T = (T_{final} - T_{initial})\)). So we can calculate the total heat transferred (\(Q\)): \(Q = (74619.45 \mathrm{kg}) \times (1.71 \mathrm{~kJ/kg\cdot K}) \times (1 \mathrm{K})\) \(Q = 127579.18 \mathrm{kJ}\) Convert it to joules: \(Q = 127579.18 \times 1000 \mathrm{J} = 127579180 \mathrm{J}\)

Determine the average rate of heat transfer.

We are given that the temperature increase occurs over a 144-hour period. To find the average rate of heat transfer, divide the total heat transferred by the time interval: \(\dot{Q}_{avg} = \frac{Q}{t} = \frac{127579180 \mathrm{J}}{144 \mathrm{hours}}\) Convert the time to seconds: \(t = 144 \times 3600 = 518400 \mathrm{s}\) Now, divide the total heat transferred by the time to find the average rate of heat transfer: \(\dot{Q}_{avg} = \frac{127579180}{518400} = 246.25 \mathrm{W}\) The closest answer to our calculated result is: (c) \(246 \mathrm{~W}\)